Find the volume of the portion of the ball r2 +y? + z2 < 2 that lies above the cone z = (Ans: (/2– 1)) nt of inertia
Find the volume of the portion of the ball r2 +y? + z2 < 2 that lies above the cone z = (Ans: (/2– 1)) nt of inertia
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Problem Statement:**
Find the volume of the portion of the ball \( x^2 + y^2 + z^2 \leq 2 \) that lies above the cone \( z = \sqrt{x^2 + y^2} \).
**Answer:**
\[
\frac{4}{3}(\sqrt{2} - 1)
\]
**Solution Explanation:**
To calculate the volume, we analyze the intersection of a sphere and a cone in three-dimensional space. The sphere is centered at the origin with a radius of \(\sqrt{2}\), and the cone has its vertex at the origin extending upwards.
To find the volume of the region above the cone but inside the sphere, one might use polar coordinates to express the boundaries of integration, performing integration in spherical or cylindrical coordinates. The symmetry of the problem can simplify the calculations, and by setting up appropriate integrals, the volume can be evaluated.
**Graphical Representation:**
A graph of this situation would typically depict a 3D coordinate system showing:
- A sphere centered at the origin with a cut-off portion by a cone.
- The cone also centered and emanating from the origin, extending upwards.
- The intersecting volume is what persists within the sphere but excluded by the cone underneath.
Further detailed calculation steps involve using triple integrals and symmetry considerations to reduce the computational complexity, leading to the volume formula provided in the answer.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc740e8b2-d410-40ba-b674-1beb1e9bb454%2Fe35df9dc-3f88-48e8-9611-1f38d7c3d910%2F61r2062_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the volume of the portion of the ball \( x^2 + y^2 + z^2 \leq 2 \) that lies above the cone \( z = \sqrt{x^2 + y^2} \).
**Answer:**
\[
\frac{4}{3}(\sqrt{2} - 1)
\]
**Solution Explanation:**
To calculate the volume, we analyze the intersection of a sphere and a cone in three-dimensional space. The sphere is centered at the origin with a radius of \(\sqrt{2}\), and the cone has its vertex at the origin extending upwards.
To find the volume of the region above the cone but inside the sphere, one might use polar coordinates to express the boundaries of integration, performing integration in spherical or cylindrical coordinates. The symmetry of the problem can simplify the calculations, and by setting up appropriate integrals, the volume can be evaluated.
**Graphical Representation:**
A graph of this situation would typically depict a 3D coordinate system showing:
- A sphere centered at the origin with a cut-off portion by a cone.
- The cone also centered and emanating from the origin, extending upwards.
- The intersecting volume is what persists within the sphere but excluded by the cone underneath.
Further detailed calculation steps involve using triple integrals and symmetry considerations to reduce the computational complexity, leading to the volume formula provided in the answer.
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